Any idea how to find at least one "n" such that:
[ ( x1/n + y1/n ) / 2 ]2n - xy < epsilon
2 <= x <= a, 2 <= y <= a
for any epsilon?
Actually, its well-known that:
lim [ ( x1/n + y1/n ) / 2 ]2n = xy
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Even for lim n^(1/n) = 1 it is a problem ...
(1/n) (log n) < log(1 + eps) , n > ?? ...
Last edited by yms; Nov 14th 2012 at 04:30 PM.
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